I am looking for a derivation of the fact that $\frac{1}{\Phi^{-1}(3/4)}$ is the multiplier needed for the Median Absolute Deviation (MAD) to be an unbiased estimator of $\sigma$ when $x_i\sim N(0, \sigma^2)$. Recall the MAD is defined as:
$$ \lambda = b\times Median(| X-Median(X)|). $$
for some $b$ chosen to meet a given criteria, (e.g. often unbiased-ness) the claim is stated in many places (wikipedia and journal articles) but I cannot find a proof.
An outline of why it is the case that $\frac{\text{MAD}}{\sigma}$ at the normal is $\Phi^{-1}(3/4)$ (where here $\text{MAD}$ is the median absolute deviation from the median).
For the normal, the population median is at $\mu$.
Then consider the distribution of $\frac{X-\mu}{\sigma}$. This is just a standard normal. Hence $\frac{|X-\mu|}{\sigma}$ is the absolute value of a standard normal. Its median will be the value $m$ such that the standard normal has half the area between $-m$ and $m$. We already know that half the distribution is between the first and third quartile, so $m$ is the upper quartile.
Consequently in a general normal distribution, the ratio $\frac{\text{MAD}}{\sigma}$ is the upper quartile, $\Phi^{-1}(3/4)$.
Now for the normal (and indeed in other symmetric distributions with finite mean with positive density at the median) the sample median is consistent as an estimator of $\mu$, so $X-\tilde{x}$ will asymptotically go to $X-\mu$, and so in turn asymptotically the median of $|X-\tilde{x}|$ will be the median of $|X-\mu|$.
As a result, to get a consistent estimator of $\sigma$ from the MAD (median absolute deviation from the median) for a random sample from a normal distribution, you divide the MAD by $\Phi^{-1}(3/4)$.
